Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
Maybe using path clipping you can plot only over land like this :
--
import numpy as np
from mpl_toolkits.basemap import Basemap
import pylab as plt
import matplotlib.patches as patches
import matplotlib.transforms as transforms
ll_lat = -38.10 # extent of area of interest
ll_lon = 145.06
ur_lat = -38.00
ur_lon = 145.16
# create data points all on land
fudge = ((ll_lon+0.05, ur_lat-0.00, 1000),
(ll_lon+0.05, ur_lat-0.01, 2000),
(ll_lon+0.05, ur_lat-0.02, 3000),
(ll_lon+0.05, ur_lat-0.03, 4000),
(ll_lon+0.04, ur_lat-0.025, 1000),
(ll_lon+0.043, ur_lat-0.036, 1),
(ll_lon+0.047, ur_lat-0.041, 2),
(ll_lon+0.07, ur_lat-0.07, 4000),
(ll_lon+0.08, ur_lat-0.08, 3000),
(ll_lon+0.09, ur_lat-0.09, 2000),
(ll_lon+0.10, ur_lat-0.10, 1000))
data = np.ones((3, len(fudge)))
for (i, (lon, lat, val)) in enumerate(fudge):
data[0,i] = lon
data[1,i] = lat
data[2,i] = val
# plot the data
fig = plt.figure()
m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat,
llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f')
# +CHANGES
coll = plt.hexbin(data[0,:], data[1,:], data[2,:], gridsize=5)
ax = plt.gca()
for n, poly in enumerate(m.coastpolygons):
type = m.coastpolygontypes[n]
if type in [1,3]:
p = patches.Polygon(np.asarray(poly).T)
p.set_color('none')
ax.add_patch(p)
m.set_axes_limits(ax=ax)
coll.set_clip_path(p)
# -CHANGES
plt.show()
--
On Wed, Nov 4, 2009 at 1:17 AM, ross.wil...@ga.gov.au wrote:
-Original Message-
From: Jeff Whitaker [mailto:jsw...@fastmail.fm]
Sent: Tuesday, 3 November 2009 02:53
To: Stephane Raynaud
Cc: Wilson Ross; matplotlib-users@lists.sourceforge.net
Subject: Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
Stephane Raynaud wrote:
Ross,
one way is to mask (or remove) ocean points using the _geoslib module
provided with basemap.
When you create a Basemap instance, you can retrieve all its polygons
land (continents and islands) with mymap.coastpolygons.
Thay are stored as numpy arrays, and you can convert them to
_geoslib.Polygon objects :
poly = _geoslib.Polygon(N.asarray(coastalpoly).T)
Then you loop over all Polygons and all (x,y) points and test :
good_point = _geoslib.Point((x,y)).within(poly)
Thanks to this method, you can choose you optimal resolution.
You can even compute the intersection of you hexagons with coastal
polygons using .intersection() and .area (instead of simply checking
if the center is inside) and then reject points depending the fraction
of the cell covered by land (or ocean).
Following Stephane's excellent suggestion, here's a prototype Basemap
method that checks to see if a point is on land or over water. Ross -
if you find it useful I'll include it in the next release. Note that it
will be slow for lots of points or large map regions.
-Jeff
--
Yes, Stephane's approach is nice and Jeff has nicely encapsulated the
approach. I'll put that into my bag of tricks!
However it doesn't quite do what I want. My data does not have any points
in the ocean; the hex binning creates hexagonal patches that extend out into
the ocean. As a physicist I say that's a representation artifact and leave
it at that, but my end-users want that 'bleed' into the ocean removed. My
argument that they are losing data falls on deaf ears.
Here's an even more contrived example that improves on my poor previous
attempt to explain the problem:
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.basemap import Basemap
ll_lat = -38.10 # extent of area of interest
ll_lon = 145.06
ur_lat = -38.00
ur_lon = 145.16
# create data points all on land
fudge = ((ll_lon+0.05, ur_lat-0.00, 1000),
(ll_lon+0.05, ur_lat-0.01, 2000),
(ll_lon+0.05, ur_lat-0.02, 3000),
(ll_lon+0.05, ur_lat-0.03, 4000),
(ll_lon+0.04, ur_lat-0.025, 1000),
(ll_lon+0.043, ur_lat-0.036, 1),
(ll_lon+0.047, ur_lat-0.041, 2),
(ll_lon+0.07, ur_lat-0.07, 4000),
(ll_lon+0.08, ur_lat-0.08, 3000),
(ll_lon+0.09, ur_lat-0.09, 2000),
(ll_lon+0.10, ur_lat-0.10, 1000))
data = np.ones((3, len(fudge)))
for (i, (lon, lat, val)) in enumerate(fudge):
data[0,i] = lon
data[1,i] = lat
data[2,i] = val
# plot the data
fig = plt.figure()
m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat,
llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f')
plt.hexbin(data[0,:], data[1,:], data[2,:], gridsize=5)
m.drawcoastlines(linewidth=0.5, color
Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
-Original Message- From: Jeff Whitaker [mailto:jsw...@fastmail.fm] Sent: Tuesday, 3 November 2009 02:53 To: Stephane Raynaud Cc: Wilson Ross; matplotlib-users@lists.sourceforge.net Subject: Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED] Stephane Raynaud wrote: Ross, one way is to mask (or remove) ocean points using the _geoslib module provided with basemap. When you create a Basemap instance, you can retrieve all its polygons land (continents and islands) with mymap.coastpolygons. Thay are stored as numpy arrays, and you can convert them to _geoslib.Polygon objects : poly = _geoslib.Polygon(N.asarray(coastalpoly).T) Then you loop over all Polygons and all (x,y) points and test : good_point = _geoslib.Point((x,y)).within(poly) Thanks to this method, you can choose you optimal resolution. You can even compute the intersection of you hexagons with coastal polygons using .intersection() and .area (instead of simply checking if the center is inside) and then reject points depending the fraction of the cell covered by land (or ocean). Following Stephane's excellent suggestion, here's a prototype Basemap method that checks to see if a point is on land or over water. Ross - if you find it useful I'll include it in the next release. Note that it will be slow for lots of points or large map regions. -Jeff -- Yes, Stephane's approach is nice and Jeff has nicely encapsulated the approach. I'll put that into my bag of tricks! However it doesn't quite do what I want. My data does not have any points in the ocean; the hex binning creates hexagonal patches that extend out into the ocean. As a physicist I say that's a representation artifact and leave it at that, but my end-users want that 'bleed' into the ocean removed. My argument that they are losing data falls on deaf ears. Here's an even more contrived example that improves on my poor previous attempt to explain the problem: import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap ll_lat = -38.10 # extent of area of interest ll_lon = 145.06 ur_lat = -38.00 ur_lon = 145.16 # create data points all on land fudge = ((ll_lon+0.05, ur_lat-0.00, 1000), (ll_lon+0.05, ur_lat-0.01, 2000), (ll_lon+0.05, ur_lat-0.02, 3000), (ll_lon+0.05, ur_lat-0.03, 4000), (ll_lon+0.04, ur_lat-0.025, 1000), (ll_lon+0.043, ur_lat-0.036, 1), (ll_lon+0.047, ur_lat-0.041, 2), (ll_lon+0.07, ur_lat-0.07, 4000), (ll_lon+0.08, ur_lat-0.08, 3000), (ll_lon+0.09, ur_lat-0.09, 2000), (ll_lon+0.10, ur_lat-0.10, 1000)) data = np.ones((3, len(fudge))) for (i, (lon, lat, val)) in enumerate(fudge): data[0,i] = lon data[1,i] = lat data[2,i] = val # plot the data fig = plt.figure() m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f') plt.hexbin(data[0,:], data[1,:], data[2,:], gridsize=5) m.drawcoastlines(linewidth=0.5, color='k') plt.show() None of the data points are on land, as can be seen if you change the 'gridsize' parameter to 100 or so (land is to the NE). However, when gridsize=5, the red hexagon in the middle extends out into the ocean. My users want that hexagon and others like it clipped at the coastline. Crazy people, users. Ross -- Let Crystal Reports handle the reporting - Free Crystal Reports 2008 30-Day trial. Simplify your report design, integration and deployment - and focus on what you do best, core application coding. Discover what's new with Crystal Reports now. http://p.sf.net/sfu/bobj-july ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
Ross, one way is to mask (or remove) ocean points using the _geoslib module provided with basemap. When you create a Basemap instance, you can retrieve all its polygons land (continents and islands) with mymap.coastpolygons. Thay are stored as numpy arrays, and you can convert them to _geoslib.Polygon objects : poly = _geoslib.Polygon(N.asarray(coastalpoly).T) Then you loop over all Polygons and all (x,y) points and test : good_point = _geoslib.Point((x,y)).within(poly) Thanks to this method, you can choose you optimal resolution. You can even compute the intersection of you hexagons with coastal polygons using .intersection() and .area (instead of simply checking if the center is inside) and then reject points depending the fraction of the cell covered by land (or ocean). On Mon, Nov 2, 2009 at 8:07 AM, ross.wil...@ga.gov.au wrote: Listers, I'm using basemap to plot randomly sampled values (x,y,z) through hexbin. This produces a very nice result. Some sample code is: -- import numpy as np from numpy.random import seed import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap from matplotlib.mlab import griddata ll_lat = -38.39477 # extent of area of interest ll_lon = 144.54767 ur_lat = -37.51642 ur_lon = 145.67144 num_points = 100# sample points # create random sampling over the area of interest seed(0) data = np.ones((3, num_points)) data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) data[2,:] *= np.random.random((num_points))*1 # plot the data fig = plt.figure() ax = fig.add_subplot(111) m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', suppress_ticks=False, area_thresh=0.5) plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) m.drawcoastlines(linewidth=0.25, color='k', zorder=2) plt.show() -- This contrived example shows a sparse set of hexagons on both land and ocean. I would like the hexagons over the ocean to be hidden. I can make the ones on land disappear by changing the 'zorder' parameter of .hexbin() to 0. However I have found no way of doing the inverse and hiding hexagons over the ocean. Using drawlsmask() is too crude at a 5-minute resolution. Any ideas? Thanks, Ross -- Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users -- Stephane Raynaud -- Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
Re: [Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
Stephane Raynaud wrote: Ross, one way is to mask (or remove) ocean points using the _geoslib module provided with basemap. When you create a Basemap instance, you can retrieve all its polygons land (continents and islands) with mymap.coastpolygons. Thay are stored as numpy arrays, and you can convert them to _geoslib.Polygon objects : poly = _geoslib.Polygon(N.asarray(coastalpoly).T) Then you loop over all Polygons and all (x,y) points and test : good_point = _geoslib.Point((x,y)).within(poly) Thanks to this method, you can choose you optimal resolution. You can even compute the intersection of you hexagons with coastal polygons using .intersection() and .area (instead of simply checking if the center is inside) and then reject points depending the fraction of the cell covered by land (or ocean). Following Stephane's excellent suggestion, here's a prototype Basemap method that checks to see if a point is on land or over water. Ross - if you find it useful I'll include it in the next release. Note that it will be slow for lots of points or large map regions. -Jeff On Mon, Nov 2, 2009 at 8:07 AM, ross.wil...@ga.gov.au mailto:ross.wil...@ga.gov.au wrote: Listers, I'm using basemap to plot randomly sampled values (x,y,z) through hexbin. This produces a very nice result. Some sample code is: -- import numpy as np from numpy.random import seed import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap from matplotlib.mlab import griddata ll_lat = -38.39477 # extent of area of interest ll_lon = 144.54767 ur_lat = -37.51642 ur_lon = 145.67144 num_points = 100# sample points # create random sampling over the area of interest seed(0) data = np.ones((3, num_points)) data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) data[2,:] *= np.random.random((num_points))*1 # plot the data fig = plt.figure() ax = fig.add_subplot(111) m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', suppress_ticks=False, area_thresh=0.5) plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) m.drawcoastlines(linewidth=0.25, color='k', zorder=2) plt.show() -- This contrived example shows a sparse set of hexagons on both land and ocean. I would like the hexagons over the ocean to be hidden. I can make the ones on land disappear by changing the 'zorder' parameter of .hexbin() to 0. However I have found no way of doing the inverse and hiding hexagons over the ocean. Using drawlsmask() is too crude at a 5-minute resolution. Any ideas? Thanks, Ross -- Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net mailto:Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users -- Stephane Raynaud -- Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users -- Jeffrey S. Whitaker Phone : (303)497-6313 Meteorologist FAX: (303)497-6449 NOAA/OAR/PSD R/PSD1Email : jeffrey.s.whita...@noaa.gov 325 BroadwayOffice : Skaggs Research Cntr 1D-113 Boulder, CO, USA 80303-3328 Web: http://tinyurl.com/5telg import numpy as np from numpy import ma from numpy.random import seed import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap, _geoslib from matplotlib.mlab import griddata ll_lat = -38.39477 # extent of area of interest ll_lon = 144.54767 ur_lat = -37.51642 ur_lon = 145.67144 num_points = 100# sample points # create random
[Matplotlib-users] basemap: Mask the ocean [SEC=UNCLASSIFIED]
Listers, I'm using basemap to plot randomly sampled values (x,y,z) through hexbin. This produces a very nice result. Some sample code is: -- import numpy as np from numpy.random import seed import matplotlib.pyplot as plt from mpl_toolkits.basemap import Basemap from matplotlib.mlab import griddata ll_lat = -38.39477 # extent of area of interest ll_lon = 144.54767 ur_lat = -37.51642 ur_lon = 145.67144 num_points = 100# sample points # create random sampling over the area of interest seed(0) data = np.ones((3, num_points)) data[0,:] *= ll_lon + np.random.random((num_points))*(ur_lon-ll_lon) data[1,:] *= ll_lat + np.random.random((num_points))*(ur_lat-ll_lat) data[2,:] *= np.random.random((num_points))*1 # plot the data fig = plt.figure() ax = fig.add_subplot(111) m = Basemap(projection='cyl', llcrnrlat=ll_lat, urcrnrlat=ur_lat, llcrnrlon=ll_lon, urcrnrlon=ur_lon, resolution='f', suppress_ticks=False, area_thresh=0.5) plt.hexbin(data[0,:], data[1,:], data[2,:], zorder=3) m.fillcontinents(color=(0.8,0.8,0.8,0), zorder=1) m.drawcoastlines(linewidth=0.25, color='k', zorder=2) plt.show() -- This contrived example shows a sparse set of hexagons on both land and ocean. I would like the hexagons over the ocean to be hidden. I can make the ones on land disappear by changing the 'zorder' parameter of .hexbin() to 0. However I have found no way of doing the inverse and hiding hexagons over the ocean. Using drawlsmask() is too crude at a 5-minute resolution. Any ideas? Thanks, Ross -- Come build with us! The BlackBerry(R) Developer Conference in SF, CA is the only developer event you need to attend this year. Jumpstart your developing skills, take BlackBerry mobile applications to market and stay ahead of the curve. Join us from November 9 - 12, 2009. Register now! http://p.sf.net/sfu/devconference ___ Matplotlib-users mailing list Matplotlib-users@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/matplotlib-users
